Find the greatest prime divisor of the value of the arithmetic series

Question

1 + 2 + 3 + \dots + 200

Answer 1

The sum of an arithmetic series can be calculated with the formula:
\[ S = \frac{n}{2}(a_1 + a_n) \]

where:
- \( S \) is the sum of the series
- \( n \) is the number of terms in the series
- \( a_1 \) is the first term in the series
- \( a_n \) is the last term in the series

In this case:
- \( n = 200 \)
- \( a_1 = 1 \)
- \( a_n = 200 \)

Plugging these values into the formula, we get:
\[ S = \frac{200}{2}(1 + 200) = 100 \times 201 = 20100 \]

Now, we need to find the greatest prime divisor of 20100. To do this, we can start by finding the prime factorization of 20100:

\[ 20100 = 2^2 \times 3^1 \times 5^2 \times 67^1 \]

The greatest prime divisor of 20100 is therefore 67.

So, the greatest prime divisor of the arithmetic series 1 + 2 + 3 + ... + 200 is 67.